In the given circuit diagram,when the current | vedclass

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(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor $C$ and resistor $r_1$.T

Vedclass · Accepted Answer

$CE \frac{r_2}{r + r_2}$

Vedclass · Answer

(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor $C$ and resistor $r_1$.The circuit reduces to a simple series circuit consisting of the battery $E$ and resistors $r$ and $r_2$.The current $i$ flowing through the circuit is given by:$i = \frac{E}{r + r_2}$The potential difference across the capacitor $V_c$ is equal to the potential difference across the resistor $r_2$ because they are connected in parallel.$V_c = i \cdot r_2 = \left( \frac{E}{r + r_2} \right) r_2$The charge $Q$ on the capacitor is given by $Q = C V_c$.Substituting the value of $V_c$:$Q = C \left( \frac{E r_2}{r + r_2} \right) = CE \frac{r_2}{r + r_2}$

In the given circuit diagram,when the current reaches a steady state,the charge on the capacitor of capacitance $C$ will be:

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